====== Question 2, Exercise 2.2 ======
Solutions of Question 2 of Exercise 2.2 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

=====Question 2(i)=====
Without evaluating state the reasons for the equalities.
$\left|\begin{matrix} 1 & 2 & 0  \\3 & 1 & 0  \\-1 & 2 & 0
\end{matrix}\right|=0$.
====Solution====
Given
$$\left| \begin{matrix}
   1 & 2 & 0  \\
   3 & 1 & 0  \\
   -1 & 2 & 0  \\
\end{matrix} \right|=0$$
Because elements of third column are zero.

=====Question 2(ii)===== 
Without evaluating state the reasons for the equalities. 
$\left| \begin{matrix}1 & 2 & 3  \\-8 & 4 & -12  \\2 & -1 & 3 \end{matrix} \right|=0$.
====Solution====
Given
$$\left| \begin{matrix}
   1 & 2 & 3  \\
   -8 & 4 & -12  \\
   2 & -1 & 3  \\
\end{matrix} \right|=0$$
Taking $-4$ common from $R_2$, we have
$$-4\left| \begin{matrix}
   1 & 2 & 3  \\
   2 & -1 & 3  \\
   2 & -1 & 3  \\
\end{matrix} \right|=0$$
As elements of second and third rows are identical, so result is zero.

=====Question 2(iii)=====
Without evaluating state the reasons for the equalities. 
$\left| \begin{matrix}1 & 3 & -2  \\3 & -1 & 1  \\2 & 1 & 4\end{matrix} \right|=\left| \begin{matrix}1 & 3 & 2  \\3 & -1 & 1  \\-2 & 1 & 4 \end{matrix} \right|$.
====Solution====
Given
$$\left| \begin{matrix}
   1 & 3 & -2  \\
   3 & -1 & 1  \\
   2 & 1 & 4  \\
\end{matrix} \right|=\left| \begin{matrix}
   1 & 3 & 2  \\
   3 & -1 & 1  \\
   -2 & 1 & 4  \\
\end{matrix} \right|$$
Right side is the transpose of left one.

=====Question 2(iv)=====
Without evaluating state the reasons for the equalities. 
$\left| \begin{matrix}3 & 2 & 0  \\1 & 1 & -3  \\2 & 4 & -6 \end{matrix} \right|=-3\left| \begin{matrix}3 & 2 & 0  \\1 & 1 & 1  \\2 & 4 & 2 \end{matrix} \right|$.
====Solution====
Given
$$\left| \begin{matrix}3 & 2 & 0  \\1 & 1 & -3  \\2 & 4 & -6\\
\end{matrix} \right|=-3\left| \begin{matrix}3 & 2 & 0  \\1 & 1 & 1  \\2 & 4 & 2\\ \end{matrix} \right|$$
Multiply $-3$ by third column of R.H.S.


=====Question 2(v)=====
Without evaluating state the reasons for the equalities. 
$\left| \begin{matrix}1 & 0 & -1  \\3 & 2 & 1  \\1 & -1 & 0 \end{matrix} \right|=-\left| \begin{matrix}1 & 0 & -1  \\1 & -1 & 0  \\3 & 2 & 1  \\\end{matrix} \right|$.
====Solution==== 
Given
$$\left| \begin{matrix}
   1 & 0 & -1  \\
   3 & 2 & 1  \\
   1 & -1 & 0  \\
\end{matrix} \right|=-\left| \begin{matrix}
   1 & 0 & -1  \\
   1 & -1 & 0  \\
   3 & 2 & 1  \\
\end{matrix} \right|$$
Interchanging the second and third rows on L.H.S.

=====Question 2(vi)=====
Without evaluating state the reasons for the equalities. 
$\left| \begin{matrix}2 & 0 & 1  \\3 & 1 & 2  \\1 & 2 & 2  \end{matrix} \right|=\left| \begin{matrix}
2 & 0 & 1  \\5 & 5 & 6  \\1 & 2 & 2 \end{matrix} \right|$.
====Solution====
Given
$$\left| \begin{matrix}
   2 & 0 & 1  \\
   3 & 1 & 2  \\
   1 & 2 & 2  \\
\end{matrix} \right|=\left| \begin{matrix}
   2 & 0 & 1  \\
   5 & 5 & 6  \\
   1 & 2 & 2  \\
\end{matrix} \right|$$
Multiply the third row by $2$ and add it in 2nd row of L.H.S to get R.H.S.


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